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\title{\heiti\zihao{2} 思考题1}
\author{中书君}
\date{\songti \today}

\begin{document}
\maketitle
\section{级数$1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\cdots$的各项重新安排,使先依次出现$p$个正项,再出现$q$个负项,如此交替,求新级数的和.}
\textbf{解}\quad
由于数列$a_{n}=\left\{\dfrac{1}{n}\right\}$单调递减,从而若其存在极限,则部分和数列的子列$S_{n(p+q)}$的极限等于原级数极限.
$$
	\begin{aligned}
		S_{n(p+q)}= & 1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2p-1}-\frac{1}{2}-\frac{1}{4}-\cdots-\frac{1}{2q}                                              \\
		            & +\frac{1}{2p+1}+\frac{1}{2p+3}+\cdots+\frac{1}{4p-1}-\frac{1}{2q+2}-\frac{1}{2q+4}-\cdots-\frac{1}{4q}                                   \\
		            & +\cdots                                                                                                                                  \\
		            & \vdots                                                                                                                                   \\
		            & +\frac{1}{2np-(2p-1)}+\frac{1}{2np-(2p-3)}+\cdots+\frac{1}{2np-1}                                                                        \\
		            & -\frac{1}{2nq-(2q-2)}-\frac{1}{2nq-(2q-4)}-\cdots-\frac{1}{2nq}                                                                          \\
		=           & 1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2np-1}-\frac{1}{2}-\frac{1}{4}-\cdots-\frac{1}{2nq}                                            \\
		=           & 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2np}-\frac{1}{2}-\frac{1}{4}-\cdots-\frac{1}{2np}-\frac{1}{2}-\frac{1}{4}-\cdots-\frac{1}{2nq} \\
		=           & \sum_{k=1}^{2np}\frac{1}{k}-\frac{1}{2}\left(\sum_{k=1}^{np}\frac{1}{k}\right)-\frac{1}{2}\left(\sum_{k=1}^{nq}\frac{1}{k}\right)
	\end{aligned}
$$
又因为
\begin{equation}
    \sum_{k=1}^{n}\frac{1}{k}=\gamma+\ln n + \varepsilon_{n}\quad(\varepsilon_{n}\rightarrow 0)
\end{equation}
由(1):
$$
\begin{aligned}
    \lim_{n\rightarrow \infty}S_{n(p+q)}&=\lim_{n\rightarrow \infty}\gamma+\ln (2np)+\varepsilon_{2np}-\frac{1}{2}\left[\gamma+\ln(np)+\varepsilon_{np}\right]-\frac{1}{2}\left[\gamma+\ln (nq)+\varepsilon_{nq}\right]\\
    &=\ln 2 +\frac{1}{2}\ln \frac{p}{q}
\end{aligned}
$$
对于思考题,$p=q$,当然最终的结果为$\ln 2$.

\section{讨论级数$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}}{n^{p}+(-1)^{n}}$的敛散性.}
\textbf{解}\quad
当$p>1$时,显然由比较判别法知:
$$
\lim_{n\rightarrow \infty}\frac{\frac{1}{n^{p}}}{\frac{1}{n^{\frac{1+p}{2}}}}=0
$$
从而级数收敛,显然也为绝对收敛.\par 
当$p\leqslant 0$时,$\lim\limits_{n\rightarrow \infty}\frac{(-1)^{n}}{n^{p}+(-1)^{n}}\neq 0$,从而级数发散.\par 
当$0<p\leqslant 1$时,考虑级数:
\begin{equation}
    \sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n^{p}+(-1)^{n}}-\frac{(-1)^{n}}{n^{p}}\right)
\end{equation}
由于
$$
\frac{(-1)^{n}}{n^{p}+(-1)^{n}}-\frac{(-1)^{n}}{n^{p}}=\frac{1}{n^{p}(n^{p}+(-1)^{n})}\sim \frac{1}{n^{2p}}
$$
所以此级数当$p>\dfrac{1}{2}$时收敛,当$0<p\leqslant\dfrac{1}{2}$时发散.\par 
又因为(2)右侧的项拆分出来是Leibniz交错级数,从而其收敛.从而(2)敛散性同于\\$\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{p}(n^{p}+(-1)^{n})}\sim \sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2p}}$.
所以此级数当$p>\dfrac{1}{2}$时收敛(显然条件收敛),当$0<p\leqslant\dfrac{1}{2}$时发散.\par 





\end{document}